How to printf a 64-bit integer as hex? [duplicate]

问题:

This question already has an answer here:

• How to print a int64_t type in C 6 answers

With the following code I am trying to output the value of a unit64_t variable using printf(). Compiling the code with gcc, returns the following warning:

warning: format ‘%x’ expects argument of type ‘unsigned int’, but argument 2 has type ‘uint64_t’ [-Wformat=]

The code:

#include <stdio.h>
#include <stdint.h>

int main ()
{
    uint64_t val = 0x1234567890abcdef;
    printf("val = 0x%x\n", val);

    return 0;
}

The output:

val = 0x90abcdef

Expected output:

val = 0x1234567890abcdef

How can I output a 64bit value as a hexadecimal integer using printf()? The x specifier seems to be wrong in this case.

回答1:

The warning from your compiler is telling you that your format specifier doesn't match the data type you're passing to it.

Try using %lx or %llx. For more portability, include inttypes.h and use the PRIx64 macro.

For example: printf("val = 0x%" PRIx64 "\n", val); (note that it's string concatenation)

回答2:

Edit: Use printf("val = 0x%" PRIx64 "\n", val); instead. Try printf("val = 0x%llx\n", val);instead. See the printf manpage:

ll (ell-ell). A following integer conversion corresponds to a long long int or unsigned long long int argument, or a following n conversion corresponds to a pointer to a long long int argument.

Edit: Even better is what @M_Oehm wrote: There is a specific macro for that, because unit64_t is not always a unsigned long long: PRIx64 see also this stackoverflow answer